Two parallel plate capacitors with fixed plates are connected to...

Let the separation between the plates of both capacitors be $d$ and the width of the plates be $b$.

The capacitance of a parallel plate capacitor with a dielectric slab partially inserted up to a length $x$ is given by:
$C = \dfrac{\varepsilon_0 b (l - x)}{d} + \dfrac{K \varepsilon_0 b x}{d}$

When the capacitor is connected to a battery of emf $V$, the force exerted on the dielectric slab is:
$F = \dfrac{1}{2} V^2 \dfrac{dC}{dx}$

Differentiating $C$ with respect to $x$, we get:
$\dfrac{dC}{dx} = \dfrac{\varepsilon_0 b (K - 1)}{d}$

Thus, the force pulling the dielectric into the capacitor is:
$F = \dfrac{\varepsilon_0 b V^2 (K - 1)}{2d}$

For the left capacitor, the force pulling the left half of the slab (with dielectric constant $K_1$) towards the left is:
$F_1 = \dfrac{\varepsilon_0 b V_1^2 (K_1 - 1)}{2d}$

For the right capacitor, the force pulling the right half of the slab (with dielectric constant $K_2$) towards the right is:
$F_2 = \dfrac{\varepsilon_0 b V_2^2 (K_2 - 1)}{2d}$

For the dielectric slab to remain in equilibrium, the net force must be zero. Therefore, the force from the left capacitor must balance the force from the right capacitor:
$F_1 = F_2$

Substituting the expressions for $F_1$ and $F_2$:
$\dfrac{\varepsilon_0 b V_1^2 (K_1 - 1)}{2d} = \dfrac{\varepsilon_0 b V_2^2 (K_2 - 1)}{2d}$

$V_1^2 (K_1 - 1) = V_2^2 (K_2 - 1)$

Rearranging to find the ratio of the emf of the left battery to that of the right battery:
$\dfrac{V_1^2}{V_2^2} = \dfrac{K_2 - 1}{K_1 - 1}$

$\dfrac{V_1}{V_2} = \sqrt{\dfrac{K_2 - 1}{K_1 - 1}}$